use rhombus abcd and the given information to find each measure.
6. ( mangle bce )
7. ( mangle bec )
8. ( ac )
9. ( mangle abd )
10. ( ad )
(there is a rhombus abcd with diagonals ac and bd intersecting at e. in triangle bec, angle at b is 59°, ec is 12 cm, and cd is 14 cm.)

Question

Accepted Answer

### Problem 6: $ m\angle BCE $
# Explanation:
## Step 1: Recall Rhombus Diagonals Property
In a rhombus, diagonals bisect the angles. Also, diagonals of a rhombus are perpendicular bisectors of each other? Wait, no, actually, diagonals of a rhombus bisect the vertex angles. Wait, looking at the diagram, triangle $ BCE $: we know $ \angle BEC $ is related? Wait, no, first, in a rhombus, all sides are equal. Wait, the given angle at $ B $ between the diagonal and side is $ 59^\circ $? Wait, the diagram shows $ \angle EBC = 59^\circ $, and $ BC = 14 $ cm, $ EC = 12 $ cm? Wait, no, $ AC $ is split into $ AE $ and $ EC $, but in a rhombus, diagonals bisect each other? Wait, no, diagonals of a rhombus bisect each other at right angles? Wait, no, diagonals of a rhombus are perpendicular bisectors? Wait, no, diagonals of a rhombus bisect the angles and bisect each other, and are perpendicular. Wait, let's correct: In a rhombus, diagonals are perpendicular (so $ \angle BEC = 90^\circ $? Wait, no, maybe I misremember. Wait, properties of a rhombus: 1. All sides are equal. 2. Diagonals bisect the angles. 3. Diagonals are perpendicular bisectors of each other (so they intersect at right angles). Wait, so $ \angle BEC = 90^\circ $? But let's check problem 6: $ m\angle BCE $. Let's look at triangle $ BCE $. We know $ BC = 14 $ cm, $ EC = 12 $ cm? Wait, no, the diagram shows $ AC $ is a diagonal, with $ EC = 12 $ cm? Wait, maybe $ AC $ is split into $ AE $ and $ EC $, but in a rhombus, diagonals bisect each other, so $ AE = EC $? Wait, no, diagonals bisect each other, so if $ EC = 12 $ cm, then $ AC = 24 $ cm (problem 8). But for problem 6: $ \angle BCE $. Let's consider triangle $ BCE $. We know $ \angle EBC = 59^\circ $, and if diagonals are perpendicular, then $ \angle BEC = 90^\circ $, so in triangle $ BCE $, angles sum to $ 180^\circ $. So $ m\angle BCE = 180^\circ - 90^\circ - 59^\circ = 31^\circ $. Wait, let's confirm:

## Step 1: Diagonals in rhombus are perpendicular.
So $ \angle BEC = 90^\circ $ (since diagonals intersect at right angles).

## Step 2: In $ \triangle BCE $, sum of angles is $ 180^\circ $.
Given $ \angle EBC = 59^\circ $, $ \angle BEC = 90^\circ $, so $ m\angle BCE = 180^\circ - 90^\circ - 59^\circ = 31^\circ $.

# Answer:
$ 31^\circ $

### Problem 7: $ m\angle BEC $
# Explanation:
## Step 1: Recall rhombus diagonals property.
In a rhombus, diagonals are perpendicular bisectors of each other, so they intersect at right angles.

## Step 2: Therefore, $ \angle BEC = 90^\circ $.

# Answer:
$ 90^\circ $

### Problem 8: $ AC $
# Explanation:
## Step 1: Diagonals of a rhombus bisect each other.
So the diagonal $ AC $ is split into two equal parts by the intersection point $ E $.

## Step 2: Given $ EC = 12 $ cm, so $ AC = 2 \times EC $.
$ AC = 2 \times 12 = 24 $ cm.

# Answer:
$ 24 $ cm

### Problem 9: $ m\angle ABD $
# Explanation:
## Step 1: Diagonals of a rhombus bisect the angles.
So diagonal $ BD $ bisects $ \angle ABC $.

## Step 2: From problem 6, $ \angle BCE = 31^\circ $, and since diagonals bisect angles, $ \angle ABD = \angle DBC = 59^\circ $? Wait, no, wait. Wait, $ \angle EBC = 59^\circ $, and since $ BD $ is a diagonal, it bisects $ \angle ABC $, but also, $ \angle ABD $ is equal to $ \angle EBC $? Wait, no, let's see: In rhombus, $ AB = BC $, and diagonals bisect angles. Wait, $ \angle ABD $ is part of $ \angle ABC $, and $ \angle EBC = 59^\circ $, but actually, since diagonals are perpendicular, and we found $ \angle BCE = 31^\circ $, but maybe $ \angle ABD = \angle BCE $? Wait, no, let's think again. Wait, in rhombus, opposite sides are parallel, and diagonals bisect the angles. Also, triangles $ ABD $ and $ CBD $ are congruent. Wait, maybe $ \angle ABD = \angle BCE $? No, wait, let's look at the angles. Wait, $ \angle ABD $: since $ BD $ is a diagonal, and $ AC $ is a diagonal, intersecting at $ E $ (right angle). $ \angle ABD $ is in triangle $ ABE $. Wait, maybe $ \angle ABD = 59^\circ $? Wait, no, the given angle is $ \angle EBC = 59^\circ $, and since $ AB = BC $ (rhombus), triangle $ ABC $ is isoceles? Wait, no, all sides are equal, so $ AB = BC = CD = DA $. Wait, maybe $ \angle ABD = 59^\circ $? Wait, no, let's check the diagram: the angle at $ B $ between $ BE $ and $ BC $ is $ 59^\circ $, so $ \angle EBC = 59^\circ $, and since $ AB = BC $, and $ BE $ is common, maybe $ \angle ABD = 59^\circ $? Wait, no, that doesn't make sense. Wait, maybe I made a mistake earlier. Wait, problem 6: $ \angle BCE = 31^\circ $, and since $ AB \parallel CD $, and diagonals bisect angles, $ \angle ABD = \angle BDC $, but maybe $ \angle ABD = 31^\circ $? No, wait, let's use the right triangle. Wait, in triangle $ ABE $, $ \angle AEB = 90^\circ $, $ \angle BAE = \angle BCE = 31^\circ $ (since diagonals bisect angles, and $ AB \parallel CD $, alternate interior angles). Wait, no, maybe $ \angle ABD = 59^\circ $? Wait, the given angle is $ 59^\circ $ at $ B $ ( $ \angle EBC = 59^\circ $ ), and since $ BD $ is a diagonal, $ \angle ABD = \angle EBC = 59^\circ $? No, that can't be, because in triangle $ BCE $, we have $ \angle EBC = 59^\circ $, $ \angle BEC = 90^\circ $, $ \angle BCE = 31^\circ $. Then, since $ AB = BC $, triangle $ ABC $ is isoceles with $ AB = BC $, so $ \angle BAC = \angle BCA = 31^\circ $, so $ \angle ABC = 180 - 31 - 31 = 118^\circ $. Then, diagonal $ BD $ bisects $ \angle ABC $, so $ \angle ABD = \frac{118^\circ}{2} = 59^\circ $. Ah, there we go. So $ \angle ABD = 59^\circ $. Wait, that matches the given angle. So $ m\angle ABD = 59^\circ $.

Wait, let's re-explain:

## Step 1: In rhombus, $ AB = BC $, so $ \triangle ABC $ is isoceles with $ AB = BC $.

## Step 2: $ \angle BAC = \angle BCA = 31^\circ $ (from problem 6, $ \angle BCE = 31^\circ $, which is $ \angle BCA $).

## Step 3: Sum of angles in $ \triangle ABC $: $ \angle ABC = 180^\circ - 31^\circ - 31^\circ = 118^\circ $.

## Step 4: Diagonal $ BD $ bisects $ \angle ABC $, so $ \angle ABD = \frac{118^\circ}{2} = 59^\circ $.

# Answer:
$ 59^\circ $

### Problem 10: $ AD $
# Explanation:
## Step 1: In a rhombus, all sides are equal, so $ AD = BC $.

## Step 2: $ BC = 14 $ cm (given in the diagram: $ CD = 14 $ cm, and in rhombus, $ BC = CD $).

## Step 3: Therefore, $ AD = 14 $ cm.

# Answer:
$ 14 $ cm

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QUESTION IMAGE

Question

Explanation:

Problem 6: \( m\angle BCE \)

Step 1: Recall Rhombus Diagonals Property

Step 1: Diagonals in rhombus are perpendicular.

Step 2: In \( \triangle BCE \), sum of angles is \( 180^\circ \).

Step 1: Recall rhombus diagonals property.

Step 2: Therefore, \( \angle BEC = 90^\circ \).

Step 1: Diagonals of a rhombus bisect each other.

Step 2: Given \( EC = 12 \) cm, so \( AC = 2 \times EC \).

Answer:

Problem 7: \( m\angle BEC \)